Problem: Simplify the following expression: $\dfrac{28t^3}{4t^5}$ You can assume $t \neq 0$.
Solution: $ \dfrac{28t^3}{4t^5} = \dfrac{28}{4} \cdot \dfrac{t^3}{t^5} $ To simplify $\frac{28}{4}$ , find the greatest common factor (GCD) of $28$ and $4$ $28 = 2 \cdot 2 \cdot 7$ $4 = 2 \cdot 2$ $ \mbox{GCD}(28, 4) = 2 \cdot 2 = 4 $ $ \dfrac{28}{4} \cdot \dfrac{t^3}{t^5} = \dfrac{4 \cdot 7}{4 \cdot 1} \cdot \dfrac{t^3}{t^5} $ $\phantom{ \dfrac{28}{4} \cdot \dfrac{3}{5}} = 7 \cdot \dfrac{t^3}{t^5} $ $ \dfrac{t^3}{t^5} = \dfrac{t \cdot t \cdot t}{t \cdot t \cdot t \cdot t \cdot t} = \dfrac{1}{t^2} $ $ 7 \cdot \dfrac{1}{t^2} = \dfrac{7}{t^2} $